What is the integral of t/(t^4+10) ?

The integral of t/(t^4+10) is 1/(2sqrt(10))arctan((t^2)/(sqrt(10)))+C

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integral of t/(t^4+10)

\int \frac{t}{t ^{4} + 10} d t

\frac{1}{2 10} arctan (\frac{t ^{2}}{10}) + C

Solution steps

\int \frac{t}{t ^{4} + 10} d t

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 10 )} d u

= \int \frac{1}{2 ( u ^{2} + 10 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 10} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{10 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{10} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{2} \cdot \frac{1}{10} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{10} arctan (\frac{t ^{2}}{10})

Simplify

\frac{1}{2} \cdot \frac{1}{10} arctan (\frac{t ^{2}}{10}) : \frac{1}{2 10} arctan (\frac{t ^{2}}{10})

= \frac{1}{2 10} arctan (\frac{t ^{2}}{10})

Add a constant to the solution

= \frac{1}{2 10} arctan (\frac{t ^{2}}{10}) + C

\frac{d}{d x} (x - y)

\int x^{3} x^{2} + 38 d x

\frac{d}{d x} (A x e^{x})

(- 1, 2), (3, - 3)

\frac{d}{d x} (2 \cdot ln (x))

What is the integral of t/(t^4+10) ?
The integral of t/(t^4+10) is 1/(2sqrt(10))arctan((t^2)/(sqrt(10)))+C